Question: In the adjoining figure, $CD$ is the diameter of a semicircle with center $O$.  Point $A$ lies on the extension of $DC$ past $C$; point $E$ lies on the semicircle, and $B$ is the point of intersection (distinct from $E$) of line segment $AE$ with the semicircle.  If length $AB$ equals length $OD$, and the measure of $\angle EOD$ is $45^\circ$, then find the measure of $\angle BAO$, in degrees.

[asy]
import graph;

unitsize(2 cm);

pair O, A, B, C, D, E;

O = (0,0);
C = (-1,0);
D = (1,0);
E = dir(45);
B = dir(165);
A = extension(B,E,C,D);

draw(arc(O,1,0,180));
draw(D--A--E--O);

label("$A$", A, W);
label("$B$", B, NW);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, NE);
label("$O$", O, S);
[/asy]
Solution: Draw $BO$.  Let $y = \angle BAO$.  Since $AB = OD = BO$, triangle $ABO$ is isosceles, so $\angle BOA = \angle BAO = y$.  Angle $\angle EBO$ is exterior to triangle $ABO$, so $\angle EBO = \angle BAO + \angle BOA = y + y = 2y$.

[asy]
import graph;

unitsize(2 cm);

pair O, A, B, C, D, E;

O = (0,0);
C = (-1,0);
D = (1,0);
E = dir(45);
B = dir(165);
A = extension(B,E,C,D);

draw(arc(O,1,0,180));
draw(D--A--E--O);
draw(B--O);

label("$A$", A, W);
label("$B$", B, NW);
label("$C$", C, S);
label("$D$", D, S);
label("$E$", E, NE);
label("$O$", O, S);
[/asy]

Triangle $BEO$ is isosceles, so $\angle BEO = \angle EBO = 2y$.  Then $\angle EOD$ is external to triangle $AEO$, so $\angle EOD = \angle EAO + \angle AEO = y + 2y = 3y$.  But $\angle EOD = 45^\circ$, so $\angle BAO = y = 45^\circ/3 = \boxed{15^\circ}$.